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I think it is a True statement, and my proof is as follows:

let $\theta $ be the additive identity in F

Given any element v in V, since v + v = $\theta$ v + $\theta$ v = 2$\theta$ v, then 2$\theta$ $\in$ F, and $\frac{\theta}{2} \in$ F, hence $\frac{\theta}{2}v \in$ V, by this argument we can construct infinite $\{2v, \frac{\theta}{2}v , 3v, \frac{\theta}{3}v...\}$

Hence V cannot be finite, and we are done.

Dinoman
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    What is the meaning of dimension of a finite set? – José Carlos Santos Mar 18 '21 at 07:43
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    Maybe you could specify the conditions on $F$ as well. Is your claim this: Let $F$ be a field and $V$ be $F$-vector space of dimension 1. Then, if $V$ is a finite set, then also $F$ is a finite set (in particular a finite field). – Steffen Plunder Mar 18 '21 at 07:51

2 Answers2

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I think your proof doesn't work like this because:

  1. For a general field, you don't necessarily have that the set $\{ 2 \theta, 3 \theta, 4 \theta, 5 \theta, ..., \theta (2 \theta )^{-1}, \theta (3 \theta)^{-1}, ... \}$ is infinite! Take $\mathbb{F}_2 = \mathbb{Z} / 2 \mathbb{Z}$ as an example...The set you use is only infinite, if the characteristic of the field is not finite.
  2. There are infinite fields with finite characteristic (see for example here or consider the algebraic closure of a finite field). Your proof only shows that the characteristic has to be finite, but that's not the same as showing that the field is a finite field.

My attempt for proving the statement would be this. (Please be doubtful, I might be wrong!)

Don't read further if you want to find a proof yourself.

Let $v \in V \setminus \{ 0 \}$ (this exists since $\mathrm{dim}(V) > 0$). We consider the map $$ M: F \to V: \lambda \mapsto \lambda v. $$

  • We show by contradiction that the map $M$ is injective. Let $\lambda, \mu \in F$ with $\lambda \neq \mu$ and $\lambda v = \mu v$, then $$ (\lambda - \mu) v = 0 $$ but $$ (\lambda - \mu)^{-1} (\lambda - \mu) v = v \neq 0 $$ which is a contradiction to the vector space axioms (a zero vector multiplied with anything has to be the zero vector again). Hence there are no such elements such that $\lambda \neq \mu$ and $\lambda v = \mu v$.

  • Since $M$ is injective, we get $$ \vert F \vert \leq \vert V \vert < \infty. $$

Steffen Plunder
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A basic fact is that for a finite dimensional vector space $V $ over a field $\mathbb F $ , we have $|V|=|\mathbb F|^{\rm {dim}V} $.

This forces $|\mathbb F|\le|V|$.

For if $\lambda\cdot v=0$ and $\lambda\ne0$, then $v=0$. Proof: $0=\lambda^{-1}\cdot0=\lambda^{-1}\cdot(\lambda\cdot v)=(\lambda^{-1}\cdot\lambda)v=1\cdot v=v$.